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Mathematical Vector Addition

Mathematical Vector Addition. Mathematical Addition of Vectors.

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Mathematical Vector Addition

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  1. Mathematical Vector Addition

  2. Mathematical Addition of Vectors The process of adding vectors can be accurately done using basic trigonometry. If you follow each step carefully, you will break down each vector into it's x and y componets and determine the magnitude and direction of the resultant vector. We have provided you with a "Vector Worksheet" to help you organize your work.

  3. 00 30 Km/hr @ 450 60 Km/hr @ 3150 900 2700 60 Km/hr @ 1350 1800 Let's say that we are adding three vectors A, B, and C. STEP #1 - Deconstruct each vector into it's X & Y components. A) To do this, you must first find the Theta angle to the x axis for each vector.

  4. 00 X component = Magnitude x cos 60 Km/hr 30 Km/hr +y 900 2700 -x y component = Magnitude x sin 60 Km/hr 1800 = 45o = 45o = 450 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle -x = (60)(cos45) = -42.4 y = (60)(sin45) = +42.4 60 Km/hr @ 3150 60 Km/hr @ 1350 30 Km/hr @ 450

  5. 00 60 Km/hr 30 Km/hr +y 900 2700 -x 60 Km/hr 1800 -42.4 +42.4

  6. 00 X component = Magnitude x cos 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y 900 2700 +x -x y component = Magnitude x sin 60 Km/hr 1800 = 45o = 45o = 450 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle -x = (60)(cos45) = -42.4 y = (60)(sin45) = +42.4 +y = (30)(sin45) = +21.2 60 Km/hr @ 3150 60 Km/hr @ 1350 30 Km/hr @ 450

  7. 00 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y 900 2700 +x -x 60 Km/hr 1800 -42.4 + 21.2 +42.4 +21.2

  8. 00 X component = Magnitude x cos 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y +x 900 2700 +x -x y component = Magnitude x sin -Y 60 Km/hr 1800 = 45o = 45o = 450 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle -x = (60)(cos45) = -42.4 +x = (60)(cos45) = +42.4 y = (60)(sin45) = +42.4 +y = (30)(sin45) = +21.2 -Y = (60)(sin45) = -42.4 60 Km/hr @ 3150 60 Km/hr @ 1350 30 Km/hr @ 450

  9. STEP #2 - List and add all x components and y components. Including all signs. These sums are the components of the resultant vector. -42.4 00 + 21.2 +42.4 +63.6 -42.4 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y +x +21.2 2700 900 +x -x -Y 60 Km/hr +42.4 +21.2 -42.4 1800 -42.4 +63.6 +21.2

  10. STEP #3 - Convert the resultant components into navigational vector notation. -42.4 00 + 21.2 +42.4 +63.6 -42.4 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y +x +21.2 2700 900 +x -x -Y 60 Km/hr +42.4 +21.2 -42.4 1800 -42.4 +63.6 +21.2 To do this, first use the Pythagorean Theorem to determine the hypotenuse. +21.2 +21.2 Resultant = √21.22 +21.22

  11. -42.4 00 + 21.2 +42.4 +63.6 -42.4 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y +x +21.2 2700 900 +x -x -Y 60 Km/hr +42.4 +21.2 -42.4 1800 -42.4 +63.6 +21.2 The magnitude of the resultant is 29.9 Km/hr. Now, find the Theta θ of the resultant using the inverse tangent formula. tan-1= l y/x l = θR 29.9

  12. Since the θR = 45o, the navigational direction of the resultant vector is NAV = 90 - θR = 45o. NAVR = 45o -42.4 00 + 21.2 +42.4 29.9 +63.6 -42.4 θR=45o 60 Km/hr 30 Km/hr +x = (30)(cos45) = +21.2 +y +y +x +21.2 2700 900 +x -x -Y 60 Km/hr +42.4 +21.2 -42.4 1800 -42.4 +63.6 29.9 Km/hr 45o +21.2

  13. Use this guide to get the navigational angle in other quadrants. Quadrant #1: NAV = 90 - θR Quadrant #2: NAV = 270 + θR Quadrant #3: NAV = 270 - θR Quadrant #4: NAV = 90 + θR Congratulations! You have successfully calculated a vector addition! The resultant vector = 29.9 Km/hr @ 45o

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